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Finding freezing point

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freezing point depression is defined as
 
ΔT=Kfbi
 
but to use that equation we need the molality of the solution which is moles of solute in kilograms of solvent. We need to calculate the molecular weight of MgCl2 to do this.
 
Mg is 24.305 amu
Cl is 35.34 amu
 
so MgCl2 24.305+2(35.34)= 94.985 g/mol and we have
     4.00 g     =0.0421119124072221929778386060957 mol
94.985 g/mol
 
and the molality is
 
4.2111x10-2 mol=0.3828355673383835725258055099609 m
   0.110 kg
 
and the finally the freezing point depression:
 
    ΔTf =(1.86)(0.38283)(2.7)= 1.9ºC
 
and the actual freezing point would be
 
0ºC - 1.9ºC = -1.9ºC
 
be careful when you read these problems to determine if they are asking for the actual freezing temperature of the solution or how much the solute changes it. In this case they are nearly the same but that's because we were working with water.